Sequence of function and convergence We can easily find example of sequence of function which is convergent but not uniformly convergent on any closed and bounded interval.
My question is: can we find a sequence of function which is convergent everywhere, but not uniformly convergent on any infinite set?
Definitely I am assuming infinite set under consideration has no isolated points.
 A: The answer is : there is no such sequences of functions.
Let's start by a lemma :
Lemma : Let $E = \left\lbrace u \in \mathbb{N}^{\mathbb{N}} : u_{n+1} \geq u_n \right\rbrace$ be the set of increasing sequences. for all $S\subset E$ with  $\text{card}(S) = 2^{\aleph_0}$, it exist a an infinite subset $B$ of $S$ such as  $\forall k \in \mathbb{N}, \: \sup \lbrace u_k : u \in B \rbrace < +\infty$
Proof : check the answer by user28111 here : Does it always exist an infinite subset of sequences that satisfy this property?
Then the main proof :
Let $(f_n)$ be a sequence of functions $\mathbb{R}\to\mathbb{R}$ such that $f_n \to f$. We define
$$B_{i,j} = \bigcap_{n=j}^{\infty}\left\lbrace x \in \mathbb{R} : |f_n(x)-f(x)| \leq \frac{1}{i} \right\rbrace$$
Remark that $B_{i,j} \subset B_{i,j+1}$
and a function $\psi : \mathbb{R} \to \mathbb{N}^{\mathbb{N}}$
$$\psi(x)_i = \inf \left\lbrace j : x\in B_{i,j} \right\rbrace$$


*

*If $\psi(\mathbb{R})$ is at most countable, $ \exists u \in \psi(\mathbb{R}), \psi^{-1}(\lbrace u \rbrace)$ is uncountable. Let's take $S= \psi^{-1}(\lbrace u \rbrace)$. 


We now show that $f_n$ is uniformly convergent on $S$ :
Let's take an arbitrary $x$ in $S$. By definition of $\psi$, $\forall x\in S, \forall i\in \mathbb{N}^*, x \in B_{i,u_i}$. Hence
$$ x \in \bigcap_{i=1}^{\infty}B_{i,u_i} $$
i.e.
$$\forall i \in \mathbb{N}^*, \exists u_i \in \mathbb{N}, \forall n \geq u_i, \forall x \in S, |f_n(x)-f(x)| \leq \frac{1}{i}$$
$f_n$ is uniformly convergent on $S$


*If $\psi(\mathbb{R})$ is uncountable, we use the lemma on $\psi(\mathbb{R})$ and we call $U$ the resulting set. Let's take $S = \phi^{-1}(U)$. Notice that $S$ is also infinite


We now show that $f_n$ is uniformly convergent on $S$ . First we define the sequence $v$ :
$$v_n = \sup \lbrace u_n : u\in U \rbrace$$
Let's take an arbitrary $x$ in $S$. we have that $\forall x\in S, \forall i\in \mathbb{N}^*, x \in B_{i,v_i}$. Indeed, as $v_i \geq \psi(x)_i$,
$$x \in B_{i,psi(x)_i} \subset B_{i,v_i}$$
Then
$$x \in \bigcap_{i=1}^{\infty} B_{i,v_i}$$
i.e.
$$\forall i \in \mathbb{N}^*, \exists v_i \in \mathbb{N}, \forall n \geq v_i, \forall x \in S, |f_n(x)-f(x)| \leq \frac{1}{i}$$
$f_n$ is uniformly convergent on $S$
